X,Y are Banach spaces, T \in L(X, Y). If T is injective,the how to prove T^-1 is a linear closed operator?

Question

I have tried to prove this as followings:

1. Im(T) is a Banach Space

2. $$T^{-1} \in L(Im(T), X)$$

3. From the Banach Inverse operator Th, I can get the $$T^{-1}$$ is a closed linear operator.

But then I start to doubt whether this process can be done, with I can't assure that the second issue holds.

If my method can be finished, then How to do it?(Or you can just ignore it). Any suggestions shall be appreciated!

Answer 1

You cannot use the theorem because the image of $T$ may not be a Banach space.

Suppose, $y_n \in Im (T)$ for all $n$, $y_n \to y$ and $T^{-1}y_n \to x$. Then we can write $y_n$ as $T(x_n)$ for some $x_n$. Hence $Tx_n \to y$. Also, $T$ is continuous and $T^{-1}y_n \to x$. Hence $y_n=T(T^{-1}y_n ) \to Tx$. Since $y$ and $Tx$ are both lim its of the same sequence $(y_n)$ we get $y=Tx$. This implies $x=T^{-1}y$ and hence the graph of $T^{-1}$ is closed.